What is the integral of (3x)/(sqrt(4-x)) ?

The integral of (3x)/(sqrt(4-x)) is -6(-1/3 (4-x)^{3/2}+4sqrt(4-x))+C

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integral of (3x)/(sqrt(4-x))

\int \frac{3 x}{4 - x} d x

- 6 (- \frac{1}{3} (4 - x)^{\frac{3}{2}} + 4 4 - x) + C

Solution steps

\int \frac{3 x}{4 - x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{4 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 4) d u

= 3 \cdot \int - 2 (- u^{2} + 4) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 (- 2 \cdot \int - u^{2} + 4 d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (- 2 (- \int u^{2} d u + \int 4 d u))

\int u^{2} d u = \frac{u ^{3}}{3}

\int 4 d u = 4 u

= 3 (- 2 (- \frac{u ^{3}}{3} + 4 u))

Substitute back

u = 4 - x

= 3 (- 2 (- \frac{( 4 - x ) ^{3}}{3} + 4 4 - x))

Simplify

3 (- 2 (- \frac{( 4 - x ) ^{3}}{3} + 4 4 - x)) : - 6 (- \frac{1}{3} (4 - x)^{\frac{3}{2}} + 4 4 - x)

= - 6 (- \frac{1}{3} (4 - x)^{\frac{3}{2}} + 4 4 - x)

Add a constant to the solution

= - 6 (- \frac{1}{3} (4 - x)^{\frac{3}{2}} + 4 4 - x) + C

\frac{10}{s ( s + 1 ) ( s + 10 )}

(- 1, - 2), (5, 4)

y \frac{d y}{d x} = x e^{- y^{2}}

\frac{5 x}{x - 2}

\int \frac{( x ^{2} - x + 12 )}{x ^{3} + 2 x} d x

What is the integral of (3x)/(sqrt(4-x)) ?
The integral of (3x)/(sqrt(4-x)) is -6(-1/3 (4-x)^{3/2}+4sqrt(4-x))+C